solve_banded — SciPy v1.15.2 Manual (original) (raw)

scipy.linalg.

scipy.linalg.solve_banded(l_and_u, ab, b, overwrite_ab=False, overwrite_b=False, check_finite=True)[source]#

Solve the equation a x = b for x, assuming a is banded matrix.

The matrix a is stored in ab using the matrix diagonal ordered form:

ab[u + i - j, j] == a[i,j]

Example of ab (shape of a is (6,6), u =1, l =2):

• a01 a12 a23 a34 a45

a00 a11 a22 a33 a44 a55 a10 a21 a32 a43 a54 * a20 a31 a42 a53 * *

Parameters:

(l, u)(integer, integer)

Number of non-zero lower and upper diagonals

ab(l + u + 1, M) array_like

Banded matrix

b(M,) or (M, K) array_like

Right-hand side

overwrite_abbool, optional

Discard data in ab (may enhance performance)

overwrite_bbool, optional

Discard data in b (may enhance performance)

check_finitebool, optional

Whether to check that the input matrices contain only finite numbers. Disabling may give a performance gain, but may result in problems (crashes, non-termination) if the inputs do contain infinities or NaNs.

Returns:

x(M,) or (M, K) ndarray

The solution to the system a x = b. Returned shape depends on the shape of b.

Examples

Solve the banded system a x = b, where:

[5  2 -1  0  0]       [0]
[1  4  2 -1  0]       [1]

a = [0 1 3 2 -1] b = [2] [0 0 1 2 2] [2] [0 0 0 1 1] [3]

There is one nonzero diagonal below the main diagonal (l = 1), and two above (u = 2). The diagonal banded form of the matrix is:

 [*  * -1 -1 -1]

ab = [* 2 2 2 2] [5 4 3 2 1] [1 1 1 1 *]

import numpy as np from scipy.linalg import solve_banded ab = np.array([[0, 0, -1, -1, -1], ... [0, 2, 2, 2, 2], ... [5, 4, 3, 2, 1], ... [1, 1, 1, 1, 0]]) b = np.array([0, 1, 2, 2, 3]) x = solve_banded((1, 2), ab, b) x array([-2.37288136, 3.93220339, -4. , 4.3559322 , -1.3559322 ])